Primitive of x cubed over x squared plus a squared squared

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Theorem

$\ds \int \frac {x^3 \rd x} {\paren {x^2 + a^2}^2} = \frac {a^2} {2 \paren {x^2 + a^2} } + \frac 1 2 \map \ln {x^2 + a^2} + C$


Proof 1

Let:

\(\ds z\) \(=\) \(\ds x^2 + a^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Derivative of Power
\(\ds \leadsto \ \ \) \(\ds \int \frac {x^3 \rd x} {\paren {x^2 + a^2}^2}\) \(=\) \(\ds \int \frac {\paren {z - a^2} } {z^2} \frac {\d z} 2\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \frac {\d z} z - \frac {a^2} 2 \int \frac {\d z} {z^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 2 \int \frac {\d z} z + \frac {a^2} {2 z} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac 1 2 \ln z + \frac {a^2} {2 z} + C\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \frac {a^2} {2 \paren {x^2 + a^2} } + \frac 1 2 \map \ln {x^2 + a^2} + C\) substituting for $z$

$\blacksquare$


Proof 2

\(\ds \int \frac {x^3 \rd x} {\paren {x^2 + a^2}^2}\) \(=\) \(\ds \int \frac {x \paren {x^2 + a^2 - a^2} } {\paren {x^2 + a^2}^2} \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {x \paren {x^2 + a^2} } {\paren {x^2 + a^2}^2} \rd x - a^2 \int \frac {x \rd x} {\paren {x^2 + a^2}^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \int \frac {x \rd x} {x^2 + a^2} - a^2 \int \frac {x \rd x} {\paren {x^2 + a^2}^2}\) simplification
\(\ds \) \(=\) \(\ds \frac 1 2 \map \ln {x^2 + a^2} - a^2 \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C\) Primitive of $\dfrac x {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac 1 2 \map \ln {x^2 + a^2} - a^2 \paren {\frac {-1} {2 \paren {x^2 - a^2} } } + C\) Primitive of $\dfrac x {\paren {x^2 + a^2}^2}$
\(\ds \) \(=\) \(\ds \frac {a^2} {2 \paren {x^2 + a^2} } + \frac 1 2 \map \ln {x^2 + a^2} + C\) simplifying

$\blacksquare$


Sources

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